sequences and series - The Lerch transcendent evaluation for the parameters HurwitzLerchPhi[z,-4 s,0]

I want to evaluate the Hurwitz zeta function

$$\Phi (z, -4s, 0)= \sum_{k=0} \frac{z^k}{k^{-4s}}$$
And $|z|<1$ and $s>1$.
I want to have un upper bound for it.
I tried even Wolfram Mathematica (to have some hint of the form if possible fot the calculus) , but without success (since I give parameters and no numbers as an input).

Answer

The OP is asking about the upper bound of the function:

$$\Phi (z, -4s, 0)= \sum_{k=1}^\infty k^{4s} z^k$$

$$|z|<1 \qquad s>1$$

First, it's obvious that:

$$\sum_{k=1}^\infty k^{4s} z^k \leq \sum_{k=1}^\infty k^{4s} |z|^k$$

So let us consider only $z>0$.

For a fixed $z$ we can see that:

$$p>q \\ \sum_{k=1}^\infty k^{4p} z^k>\sum_{k=1}^\infty k^{4q} z^k$$

Which means that if an upper bound exists for $\Phi$ as a function of $s$, it will be the limit:

$$\lim_{s \to +\infty} \sum_{k=1}^\infty k^{4s} z^k = \infty$$

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Now let us fix a finite $s$ and see what happens for $z \to 1$. For $z=1$ the series obviously diverges, which automatically means that for $z$ close to $1$ the value can be as large as we want, which means there's no upper bound for a fixed $s$ either.

More rigorously, we need to prove that for any $N>0$ there exists $\epsilon >0$ such that:

$$\sum_{k=1}^\infty k^{4s} (1-\epsilon)^k > N$$

It's rather easy, we can just compare to geometric series:

$$\sum_{k=1}^\infty k^{4s} (1-\epsilon)^k>\sum_{k=1}^\infty (1-\epsilon)^k=\frac{1-\epsilon}{\epsilon}=\frac{1}{\epsilon}-1$$

Now pick $\epsilon=\frac{1}{N+1}$ and the proof is finished.